'In Perl, how can I generate random strings consisting of eight hex digits?

Using Perl, without using any extra modules that don't come with ActivePerl, how can I create a string of 8 characters from 0-F. example 0F1672DA? The padding should be controllable and exactly 8 characters is preferable.

More examples of the kinds of strings I would like to generate:

28DA9782
55C7128A
stringrandomactiveperlperl


Solution 1:[1]

General example, allowing any range of characters:

my @chars = ('0'..'9', 'A'..'F');
my $len = 8;
my $string;
while($len--){ $string .= $chars[rand @chars] };
print "$string\n";

Solution 2:[2]

sprintf("%08X", rand(0xFFFFFFFF))

some people mentioned the windows-limit of rand with the MAX-Value of rand(0x7FFF) or rand(32768) decimal, i would overcome this with binary shifting-operator '<<'

# overcomes the windows-rand()-only-works-with-max-15bit-(32767)-limitation:
#   needed 8*4==32bit random-number:
#     first get the 15 high-significant bits shift them 17bits to the left,
#     then the next 15bits shifted 2 bits to the left,
#     then the last 2 bits with no shifting:
printf( '%08X', (
    (rand(0x8000)<<17) + (rand(0x8000)<<2) + rand(0b100) )
      );

But i consider this only as academic, because it is really awkward code which is difficult to understand.
I would not use this in real-life code, only if speed mater the most.
But maybe its the fastest solution and it's demonstrating a schema to overcome the limitation of the rand()-function under windows...

Solution 3:[3]

Use sprintf to convert numbers to hex.

$foo .= sprintf("%x", rand 16) for 1..8;

Solution 4:[4]

From the "golf" comment by daxim:

perl -e'print[0..9,A..F]->[rand 16]for 1..8'

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1 Oleg V. Volkov
Solution 2
Solution 3 Sinan Ünür
Solution 4

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